Probability Problem

1. The problem statement The United States Senate contains two senators from each of the 50 states. (a) If a committee of eight senators is selected at random, what is the probability that it will contain at least one of the two senators from a certain speciﬁed state? (b) What is the probability that a group of 50 senators selected at random will contain one senator from each state?

2. Relevant equations
Combinatorial Methods

3. The attempt at a solution
I see this problem like trying to determine the probability that at least one senator from California will be in the committee, so the number of ways how this can happen is 1 or 2 senators are in the committee, so it would be:

Assuming that one senator is already in the committee then what is left to do is to place the remaining 7 from the 99 available senators. This would the number of outcomes in the case in which there is just one senator. And assuming in the next case that there are 2 senators from the same state then we just place other 6 senators from the 98 available senators.

P(N)= ( (99 choose 7)+(98 choose 6))/100 choose 8

For part B, I imagine this like making little groups of 1 senators from a set of 2 senators from each state, so the number of outcomes would 2^50 and the sample space would be the same 100 choose 8, so the probability is the ratio of what was mentioned.

How can I increase my ability to detect mistakes?. When I do a math problem, I cant see my mistake and then I look at the solution and realized that it was a very stupid mistake which was very easy to see. I can do and study my hw, but what if I try every problem without getting the right answer?. I check the answer key and realize that it was easy and I feel like I understand it, but then it keeps happening all the time. How can I be more accurate in knowledge?

I triggered on your last paragraph. If you find a foolproof answer to this question, please let me know. And I guess the rest of the world would also be very interested !

In the mean time I can offer a few tips:
1. try to turn things around. In as many ways as you can think of.

For example, check if your exercise become easier if you ask: what are the chances there is NO Ca senator in a random 8 person committee.

Another example, also having to do with turning things around: ask yourself what the composer of the exercise might have had in mind. Usually an exercise is related to a section of the book, to a series of lectures, or whatever. What you need to solve the problem is probably in that material. And if you turn that around once more: if, after some work, you end up with a seventh degree polynomial to solve, you probably are on the wrong track and should try something else. As you say: most exercises are easy,

Yet another example: don't think that you can't do it. Think I can and I will. Your nickname should be MathWiz, not MathNoob.

For part A, like BvU said, its easier to say that the probability of having at least one is equal to one minus the probability of there being none.
So, for a committee, like you said, there are 100 choose 8 ways to fill it. There are 98 choose 8 ways to fill the committee using zero from a particular state.

For part B, there are 100 choose 50 ways to choose 50 senators. As you said, there are 2^50 ways to choose one senator from each state.
I got something on the order of 10^{-14}.

As far as error checking, I often do it the hard way one time, and build a table of possibilities. If the answer is correct, then my probabilities for all possibilities should sum to one. This is pretty vague, and I often spend more time checking my work than I do actually doing my work.
Also, there are good tools for simple problems built into excel or matlab, so work a small example to test your method and then apply it with confidence to the larger problem.

I triggered on your last paragraph. If you find a foolproof answer to this question, please let me know. And I guess the rest of the world would also be very interested !

In the mean time I can offer a few tips:
1. try to turn things around. In as many ways as you can think of.

For example, check if your exercise become easier if you ask: what are the chances there is NO Ca senator in a random 8 person committee.

Another example, also having to do with turning things around: ask yourself what the composer of the exercise might have had in mind. Usually an exercise is related to a section of the book, to a series of lectures, or whatever. What you need to solve the problem is probably in that material. And if you turn that around once more: if, after some work, you end up with a seventh degree polynomial to solve, you probably are on the wrong track and should try something else. As you say: most exercises are easy,

Yet another example: don't think that you can't do it. Think I can and I will. Your nickname should be MathWiz, not MathNoob.

Hi BvU, thank you so much for your answer , but why my answer doesnt work?. I realized that I have to multiply (99 choose 7) times 2 because I did not consider the case where the other senator is put, but now I got 0.16 instead of 0.15 which is the right answer based on your method.

1. The problem statement The United States Senate contains two senators from each of the 50 states. (a) If a committee of eight senators is selected at random, what is the probability that it will contain at least one of the two senators from a certain speciﬁed state? (b) What is the probability that a group of 50 senators selected at random will contain one senator from each state?

2. Relevant equations
Combinatorial Methods

3. The attempt at a solution
I see this problem like trying to determine the probability that at least one senator from California will be in the committee, so the number of ways how this can happen is 1 or 2 senators are in the committee, so it would be:

Assuming that one senator is already in the committee then what is left to do is to place the remaining 7 from the 99 available senators. This would the number of outcomes in the case in which there is just one senator. And assuming in the next case that there are 2 senators from the same state then we just place other 6 senators from the 98 available senators.

P(N)= ( (99 choose 7)+(98 choose 6))/100 choose 8

For part B, I imagine this like making little groups of 1 senators from a set of 2 senators from each state, so the number of outcomes would 2^50 and the sample space would be the same 100 choose 8, so the probability is the ratio of what was mentioned.

How can I increase my ability to detect mistakes?. When I do a math problem, I cant see my mistake and then I look at the solution and realized that it was a very stupid mistake which was very easy to see. I can do and study my hw, but what if I try every problem without getting the right answer?. I check the answer key and realize that it was easy and I feel like I understand it, but then it keeps happening all the time. How can I be more accurate in knowledge?

I'll use notation C(a,b) for (a choose b). You claim the number of committees containing >= 1 Californian is C(99,7) + C(98,6). Well, C(99,7) is the number of ways of choosing 7 senators from a group of 99 senators, but who is to say those 99 have 98 non-Californians and 1 Californian, and who is to say that a Californian is not among the 7 chosen (which would put 2 Californians on the committee)? You need to look as well at the one not contained in the 99.

In fact, you can look at this as a "hypergeometric" distribution problem, involving a population of size 100, with two of type I (Californians) and 98 of type II (non-Californians). The probability of having exactly one Californian in the committee is the hypergeometric probability
[tex] p(1) = H(1|2,98;8) = \frac{C(2,1) \, C(98,7)}{C(100,8)} [/tex]
while the probability of having two Californians in the committee is
[tex] p(2) = H(2|2,98;8) = \frac{C(2,2) \, C(98,6)}{C(100,8)} .[/tex]
However, as already mentioned, it is easier to use ##P(\geq 1 \: \text{Califonian}) = 1 - P(0\: \text{Californians})##.

I'll use notation C(a,b) for (a choose b). You claim the number of committees containing >= 1 Californian is C(99,7) + C(98,6). Well, C(99,7) is the number of ways of choosing 7 senators from a group of 99 senators, but who is to say those 99 have 98 non-Californians and 1 Californian, and who is to say that a Californian is not among the 7 chosen (which would put 2 Californians on the committee)? You need to look as well at the one not contained in the 99.

In fact, you can look at this as a "hypergeometric" distribution problem, involving a population of size 100, with two of type I (Californians) and 98 of type II (non-Californians). The probability of having exactly one Californian in the committee is the hypergeometric probability
[tex] p(1) = H(1|2,98;8) = \frac{C(2,1) \, C(98,7)}{C(100,8)} [/tex]
while the probability of having two Californians in the committee is
[tex] p(2) = H(2|2,98;8) = \frac{C(2,2) \, C(98,6)}{C(100,8)} .[/tex]
However, as already mentioned, it is easier to use ##P(\geq 1 \: \text{Califonian}) = 1 - P(0\: \text{Californians})##.

It is certainly useful to develop a repertoire of cross-checks, but they depend on the type of problem. For these discrete maths problems, a good one is to substitute much smaller numbers and see if the same logic produces the right answer. Two states, for example. Of course, you have to be alert to where the logic might break down in order to avoid picking numbers so small they don't pick up the error.
In other types of problems, check extreme / asymptotic values, dimensional consistency, symmetry preservation...